Form5 add maths project 2009

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Form5 add maths project 2009

Postby kopipeng on Sun Apr 26, 2009 5:57 pm

as title above,anyone can giv me the answer about the form 5 add math project 2009?? tis year topic is about circle.. anyone can help me out?
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Re: Form5 add maths project 2009

Postby Lukhman al-Yusufi on Sun Apr 26, 2009 9:30 pm

For Part 3, I only managed to solve questions (a) and (b).

Question (a)
Image

Question (b)
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Re: Form5 add maths project 2009

Postby kopipeng on Sun Apr 26, 2009 10:49 pm

how about other part?? anyone can share wif me?
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Re: Form5 add maths project 2009

Postby skkwee on Tue Apr 28, 2009 9:24 pm

First of all, let's include the question of the project for the convenient of discussion.

PART 1

There are a lot of things around us related to circles or parts of a circle.

(a) Collect pictures of 5 such objects. You may use camera to take pictures around your school compound or get pictures from magazines, newspapers, the internet or any other resources.

(b) Pi or π is a mathematical constant related to circles. Define π and write a brief history of π.

PART 2

(a) Diagram 1 shows a semicircle PQR of diameter 10 cm. Semicircles PAB and BCR of diameter d1 and d2 respectively are inscribed in the semicircle PQR such that the sum of d1 and d2 is equal to 10 cm.

circles-01.png
circles-01.png (7.06 KiB) Viewed 162983 times


Complete Table 1 by using various values of d1 and the corresponding values of d2. Hence, determine the relation between the lengths of arcs PQR, PAB and BCR.

circles-02.png
circles-02.png (23.13 KiB) Viewed 161860 times


(b) Diagram 2 shows a semicircle PQR of diameter 10 cm. Semicircles PAB, BCD and DER of diameter d1, d2 and d3 respectively are inscribed in the semicircle PQR such that the sum of d1, d2 and d3 is equal to 10 cm.

circles-03.png
circles-03.png (6.86 KiB) Viewed 162425 times


(i) Using various values of d1 and d2 and the corresponding values of d3, determine the relation between the lengths of arcs PQR, PAB, BCD and DER. Tabulate your findings.

(ii) Based on your findings in (a) and (b), make generalisations about the length of the arc of the outer semicircle and the lengths of arcs of the inner semicircles for n inner semicircles where n = 2, 3, 4....

(c) For different values of diameters of the outer semicircle, show that the generalisations stated in b (ii) is still true.

PART 3

The Mathematics Society is given a task to design a garden to beautify the school by using the design as shown in Diagram 3. The shaded region will be planted with flowers and the two inner semicircles are fish ponds.

circles-04.png
circles-04.png (11.65 KiB) Viewed 162552 times


(a) The area of the flower plot is y m2 and the diameter of one of the fish ponds is x m. Express y in terms of it and x.

(b) Find the diameters of the two fish ponds if the area of the flower plot is 16.5 m2. (Use π=22/7 )

(c) Reduce the non-linear equation obtained in (a) to simple linear form and hence, plot a straight line graph. Using the straight line graph, determine the area of the flower plot if the diameter of one of the fish ponds is 4.5 m.

(d) The cost of constructing the fish ponds is higher than that of the flower plot. Use two methods to determine the area of the flower plot such that the cost of constructing the garden is minimum.

(e) The principal suggested an additional of 12 semicircular flower beds to the design submitted by the Mathematics Society as shown in Diagram 4. The sum of the diameters of the semicircular flower beds is 10 m.

circles-05.png
circles-05.png (10.37 KiB) Viewed 162290 times


The diameter of the smallest flower bed is 30 cm and the diameter of the flower beds are increased by a constant value successively. Determine the diameter of the remaining flower beds.
Image
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Re: Form5 add maths project 2009

Postby skkwee on Wed Apr 29, 2009 12:12 am

Well, this year's form 5 add math project is really easy. Much easier than previous years. Therefore, you are encouraged to do it by yourself, and then compare your answer with the proposed solution here, or discuss your problem with our friends here.

The proposed solution is NOT mean for you to copy. It just serve as a reference, beside being a platform for all students to discuss their work.

PART 1

(a) Collect pictures of 5 such objects. You may use camera to take pictures around your school compound or get pictures from magazines, newspapers, the internet or any other resources.

Well, I leave this to your own self. There thousands of object which is round in shape.

(b) Pi or π is a mathematical constant related to circles. Define π and write a brief history of π.

There are many websites providing information of pi. So, just google it. Below are some websites I found through google. I think it's more than enough for you to write a thesis. ;)

Definition: Ratio of circle's circumference to its diameter

Video
http://video.google.com/videoplay?docid ... 4291031420

Encyclopedia
http://en.wikipedia.org/wiki/Pi
http://reference.allrefer.com/encyclopedia/P/pi.html
http://www.bartleby.com/65/pi/pi.html
http://www.britannica.com/EBchecked/topic/458986/pi
http://encarta.msn.com/encyclopedia_761552884/Pi.html

Webpage
http://scienceblogs.com/goodmath/2006/08/post_2.php
http://mathforum.org/dr.math/faq/faq.pi.html
http://www.gap-system.org/~history/Hist ... _ages.html

PART 2
(a) Diagram 1 shows a semicircle PQR of diameter 10 cm. Semicircles PAB and BCR of diameter d1 and d2 respectively are inscribed in the semicircle PQR such that the sum of d1 and d2 is equal to 10 cm.

Complete Table 1 by using various values of d1 and the corresponding values of d2. Hence, determine the relation between the lengths of arcs PQR, PAB and BCR.


The length of arc (s) of a circle can be found by using the formula
{\rm{s}} = \frac{1}{2}(2\pi r) = \pi r


where r is the radius.

The result is as below:
table-1.png
table-1.png (31.58 KiB) Viewed 167481 times


From the table, we can conclude that

Length of arc PQR = Length of arc PAB + Length of arc BCR

(b) Diagram 2 shows a semicircle PQR of diameter 10 cm. Semicircles PAB, BCD and DER of diameter d1, d2 and d3 respectively are inscribed in the semicircle PQR such that the sum of d1, d2 and d3 is equal to 10 cm.

(i) Using various values of d1 and d2 and the corresponding values of d3, determine the relation between the lengths of arcs PQR, PAB, BCD and DER. Tabulate your findings.


Again, we use the same formula to find the length of arc of PQR, PAB, BCD and DER.

{\rm{s}} = \frac{1}{2}(2\pi r) = \pi r


This time, the table is a big one. You can click on the table to change to the full size view.

table-2.png
table-2.png (52.46 KiB) Viewed 167392 times


Again, we can conclude that:


Length of arc PQR = Length of arc PAB + Length of arc BCD + Length of arc CDR

(ii) Based on your findings in (a) and (b), make generalisations about the length of the arc of the outer semicircle and the lengths of arcs of the inner semicircles for n inner semicircles where n = 2, 3, 4....

Base on the findings in the table in(a) and (b) above, we conclude that:

The length of the arc of the outer semicircle is equal to the sum of the length of arcs of any number of the inner semicircles.

(c) For different values of diameters of the outer semicircle, show that the generalisations stated in b (ii) is still true.

circle6.png
circle6.png (10.23 KiB) Viewed 166997 times


Diagram above shows a big semicircle with n number of small inner circle. From the diagram, we can see that

d = d_1  + d_2  + ........... + d_n


The length of arc of the outer semicircle
s_{out}  = \frac{1}{2}(2\pi r) = \frac{1}{2}(2\pi \frac{d}{2}) = \frac{{\pi d}}{2}


The sum of the length of arcs of the inner semicircles
{\mathop{\rm s}\nolimits} _{in}  = {{\pi d_1 } \over 2} + {{\pi d_2 } \over 2} + .............. + {{\pi d_n } \over 2}


Factorise π/2
{\mathop{\rm s}\nolimits} _{in}  = {\pi  \over 2}(d_1  + d_2  + ........... + d_n )


Substitute
d_1  + d_2  + ........... + d_n = d


We get,
{\mathop{\rm s}\nolimits} _{in}  = {\pi  \over 2}(d) = {{\pi d} \over 2}


where d is any positive real number.

We can see that
{\mathop{\rm s}\nolimits} _{in}  = {\mathop{\rm s}\nolimits} _{out}


As a result, we can conclude that
The length of the arc of the outer semicircle is equal to the sum of the length of arcs of any number of the inner semicircles. This is true for any value of the diameter of the semicircle.

In other words, for different values of diameters of the outer semicircle, show that the generalisations stated in b (ii) is still true.


To be CONTINUE ...................
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Re: Form5 add maths project 2009

Postby cschuah92 on Wed Apr 29, 2009 1:40 am

Since post 6 had answered Part 1 and Part 2, I will just show the solution for Part 3.
As in post 2 had answered question 3 (a) and 3 (b), I will briefly explain question 3 (c), 3 (d) and 3 (e).
Part 3 is a little bit mind twisting.

3 (c) Linear Law
y = -πx^2/4 + 5πx/2
Change it to linear form of Y = mX + C.
y/x = -πx/4 + 5π/2
Y = y/x
X = x
m = -π/4
C = 5π/2
Thus, plot a graph of y/x against x and draw the line of best fit.
Find the value of y/x when x = 4.5 m.
Then multiply y/x you get with 4.5 to get the actual value of y.

3 (d) We need to get the largest value of y so that the cost of constructing the garden is minimum.
Method 1: Differentiation
y = -πx^2/4 + 5πx/2
dy/dx = -πx/2 + 5π/2
(d^2)y/dx^2 = -π/2 <--- y has a maximum value.
At maximum point, (d^2)y/dx^2 = 0.
-πx/2 + 5π/2 = 0
πx/2 = 5π/2
x = 5 m
x = 5 m:
maximum value of y = -π(5^2)/4 + 5π(5)/2
= 6.25π m^2
Method 2: Completing the Square
y = -πx^2/4 + 5πx/2
= -π/4(x^2 - 10x)
= -π/4(x^2 - 10x + 25 - 25)
= -π/4[(x - 5)^2 - 25]
= -π/4(x - 5)^2 + 25π/4
y is a n shape graph as a = -π/4.
Hence, it has a maximum value.
When x = 5 m, maximum value of the graph = 6.25π m^2.

3 (e) Arithmetic Progression
The keywords are:
The principal suggested an additional of 12 semicircular flower beds to the design submitted by the Mathematics Society. (n = 12)
The sum of the diameters of the semicircular flower beds is 10 m. (S12 = 10 m)
The diameter of the smallest flower bed is 30 cm. (a = 30 cm = 0.3 m)
The diameter of the flower beds are increased by a constant value successively. (d = ?)
S12 = (n/2)[2a + (n - 1)d]
10 = (12/2)[2(0.3) + (12-1)d]
= 6(0.6 + 11d)
= 3.6 + 66d
66d = 6.4
d = 16/165
Since the first flower bed is 0.3 m,
Hence the diameters of remaining 11 flower beds expressed in arithmetic progression are:
131/330 m, 163/330 m, 13/22 m, 227/330 m, 259/330 m, 97/110 m, 323/330 m, 71/66 m, 129/110 m, 419/330 m, 41/30 m

If there is any mistake please kindly PM or e-mail me.
I haven't done my project work yet but this is my brief working on it.
Let me know my mistake so that I won't get wrong when I started doing it.

To post 6: (This is what my teacher said.)
Part 2 (b) (ii):
You need to make a generalization by induction for findings in (a) and (b) (ii) but not a statement.
I suggest,
(s out) = ∑ n (s in), n = 2, 3, 4, ......
where,
s in = length of arc of inner semicircle
s out = length of arc of outer semicircle
Part 2 (c):
You need to take at least 2 different values of diameters for the outer semicircle.
You need to show your tables for each value of those diameters.
This part is only to prove that your generalization stated in Part 2 (b) (ii) is still true.
But it is good to make a proof for it. :D
Last edited by cschuah92 on Fri May 01, 2009 5:59 pm, edited 2 times in total.
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Re: Form5 add maths project 2009

Postby wow1wow2 on Wed Apr 29, 2009 9:17 pm

now i can sit back and copy paste, anyone know how to clear the one school banner behind the graph? reluctant to do so, anyway project is shty ...
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Re: Form5 add maths project 2009

Postby cschuah92 on Wed Apr 29, 2009 10:26 pm

LOL!
Don't just copy and paste...
Make your mind work...
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Re: Form5 add maths project 2009

Postby Jasmine2me on Fri May 01, 2009 12:39 am

[quote="cschuah92"]Since post 6 had answered Part 1 and Part 2, I will just show the solution for Part 3.
As in post 2 had answered question 3 (a) and 3 (b), I will briefly explain question 3 (c), 3 (d) and 3 (e).
Part 3 is a little bit mind twisting.

3 (c) Linear Law
y = -πx^2/4 + 5πx/2
Change it to linear form of Y = mX + C.
y/x = -πx/4 + 5π/2
Y = y/x
X = x
m = -π/4
C = 5π/2
Thus, plot a graph of y/x against x and draw the line of best fit.
Find the value of y/x when x = 4.5 m.
Then multiply y/x you get with 4.5 to get the actual value of y.

3 (e) Arithmetic Progression
a = 30 cm
= 0.3 m
S12 = 10 m
n = 12
S12 = (n/2)[2a + (n - 1)d]
10 = (12/2)[2(0.3) + (12-1)d]
= 6(0.6 + 11d)
= 3.6 + 66d
66d = 6.4
d = 16/165
Sequence of arithmetic progression:
3/10, 131/330, 163/330, 13/22, 227/330, 259/330, 97/110, 323/330, 71/66, 129/110, 419/330, 41/30


If there is any mistake please kindly PM or e-mail me.
I haven't done my project work yet but this is my brief working on it.
Let me know my mistake so that I won't get wrong when I started doing it.

...
...
your informations helped alot.. but the question for 3(e) asked for the diameter of the remaining flower beds
I dont understand how the A.P became the answer......as in .. what about the remaining flower beds part?
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Re: Form5 add maths project 2009

Postby Cretos92 on Fri May 01, 2009 12:42 am

PLS show me the graph!!I seriously dun ni\o how to plot in using the computer
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